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Help finding the area of the trapezoid and pythagorean th. may be needed!

Help finding the area of the trapezoid and pythagorean th. may be needed!-example-1

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Given:

The length of the side a is, 11.

The length of the other side b is, a+4 = 11+4 = 15.

The objective is to find the area of trapezium.

The formula to find the area of trapezium is,


A=(1)/(2)h(a+b)

Let's find the height of the trapezium using Pythagoras theorem.


\begin{gathered} \text{Hypotenuse}^2=Opposite^2+Adjacent^2 \\ 5^2=h^2+4^2 \\ h^2=5^2-4^2 \\ h^2=25-16 \\ h^2=9 \\ h=\sqrt[]{9} \\ h=3 \end{gathered}

Now, substitute the obtianed values in the formula to find the area of trapezium.


\begin{gathered} A=(1)/(2)\cdot3(11+15) \\ A=(3)/(2)(26) \\ A=39 \end{gathered}

Hence, the area of trapezium is 39 square units.

User Tuxmania
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